An interesting question about the curl of a vector field. Suppose we have a smooth surface in 3D, called $S$. ${\bf n}$ is the unit normal vector. Suppose locally, we have curvilinear coordinates $s,t$ such that ${\bf s}={\bf r}_s$, ${\bf t}={\bf r}_t$ and $\{{\bf s},{\bf t},{\bf n}\}$ forms an orthonormal basis. Without loss of generality, assume ${\bf r}(0,0)={\bf 0}$.
Consider a smooth extension of ${\bf n}$ into a ball $B({\bf 0},\epsilon)$ for some $\epsilon>0$ and $|{\bf n}|=1$. One possible example is ${\bf n}=\nabla\varphi/|\nabla\varphi|$ where $\varphi$ is the signed distance function.
$({\bf t}\cdot\nabla{\bf n})\cdot{\bf s}$ and $({\bf s}\cdot\nabla{\bf n})\cdot{\bf t}$(evaluated at $s=0,t=0$) are independent of the extension, where ${\bf t}\cdot\nabla{\bf n}$ is defined to be $t_i\partial_i{\bf n}$ with Einstein summation convention used.
Actually, one can simply consider ${\bf t}\cdot\nabla n_1=\frac{\partial}{\partial t}n_1({\bf r}(s,t))|_{s=0,t=0}$. This is only determined by the values of $n_1$ on the surface. Then, ${\bf t}\cdot\nabla{\bf n}$ is independent of the extension.
Then, let's consider $g=({\bf t}\cdot\nabla{\bf n})\cdot{\bf s}-({\bf s}\cdot\nabla{\bf n})\cdot{\bf t}=[{\bf t}\cdot(\nabla{\bf n}-\nabla{\bf n}^T)]\cdot{\bf s}
=-{\bf t}\cdot[(\nabla\times{\bf n})\times{\bf s}]=-(\nabla\times{\bf n})\cdot{\bf n}$ at ${\bf 0}$, which should be independent of the extension as well.
Let's use the extension ${\bf n}=\nabla\varphi/|\nabla\varphi|$. Then, $\nabla\times{\bf n}=\nabla(1/|\nabla\varphi|)\times\nabla\varphi$ which is perpendicular with ${\bf n}$. This means $g=0$ or ${\bf n}\cdot(\nabla\times{\bf n})=0$ on the surface for any extension.
However, consider ${\bf v}=(x-z,y-z,-2z)$. At point, $(1,0,0)$, in a neighborhood, we can find a smooth surface such that ${\bf v}/|{\bf v}|$ is the normal vector of that surface. This seems to suggest ${\bf v}\cdot(\nabla\times({\bf v}/|{\bf v}|))=0$ at $(1,0,0)$. This means
$\frac{1}{|{\bf v}|}{\bf v}\cdot(\nabla\times{\bf v})=0$. One can verify directly that this is not true. Where does the argument fail?
 A: I believe the answer, as you already indicated in your comment, is that your assumption "we can find a smooth surface such that $\mathbf v/|\mathbf v|$ is the normal vector of that surface" is wrong and you've found a counterexample for it. A somewhat simpler counterexample that might give you a feel for why this is not true is afforded by the vector field $\mathbf v=(z,y,0)$ with $\mathbf v\cdot(\nabla\times\mathbf v)=y\ne\mathbf0$ for $\mathbf v\ne\mathbf 0$. The normal has no $z$ component anywhere, so lines parallel to the $z$ axis always lie in the surface (so it must be a ruled surface). If you consider this at a point on the $z$ axis, the $z$ axis must be part of the surface, and the $v_y=y$ component would force the surface to curve like a cylinder, but that's incompatible with the changing $v_x=z$ component. I know that's a rather handwavy argument, I find it hard to visualize such things, but since you've already proved the impossibility, perhaps this example can assist you in believing in your proof :-).
